Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $x = \dfrac{9z + 90}{z^2 + 17z + 72} \times \dfrac{-4z - 32}{-5z - 50} $
First factor the quadratic. $x = \dfrac{9z + 90}{(z + 8)(z + 9)} \times \dfrac{-4z - 32}{-5z - 50} $ Then factor out any other terms. $x = \dfrac{9(z + 10)}{(z + 8)(z + 9)} \times \dfrac{-4(z + 8)}{-5(z + 10)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac{ 9(z + 10) \times -4(z + 8) } { (z + 8)(z + 9) \times -5(z + 10) } $ $x = \dfrac{ -36(z + 10)(z + 8)}{ -5(z + 8)(z + 9)(z + 10)} $ Notice that $(z + 10)$ and $(z + 8)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac{ -36(z + 10)\cancel{(z + 8)}}{ -5\cancel{(z + 8)}(z + 9)(z + 10)} $ We are dividing by $z + 8$ , so $z + 8 \neq 0$ Therefore, $z \neq -8$ $x = \dfrac{ -36\cancel{(z + 10)}\cancel{(z + 8)}}{ -5\cancel{(z + 8)}(z + 9)\cancel{(z + 10)}} $ We are dividing by $z + 10$ , so $z + 10 \neq 0$ Therefore, $z \neq -10$ $x = \dfrac{-36}{-5(z + 9)} $ $x = \dfrac{36}{5(z + 9)} ; \space z \neq -8 ; \space z \neq -10 $